why $\frac{(1^T \cdot x)^2}{\min(2,\sqrt{x_3})}$ is a convex function where $x \in \mathbb R^n (n>3)$?

Question

I recently watched a YouTube video titled "Stanford EE364/A Convex Optimization Stephen Boyd I 2023 I Lecture 4." I'm having trouble understanding one of the examples Professor Boyd discussed regarding preserving convexity. Specifically, it pertains to the function $\frac{z^2}{y}$ where $y \gt 0$ and $z \in \mathbb R$, which is a known convex function.

However, Professor Boyd mentioned that we can replace z with an affine function, for example, $1^T \cdot x$, where $x \in \mathbb R^n (n>3)$. Additionally, we can replace y with $\min(2,\sqrt{x_3})$, which is a concave function. Professor Boyd explained that this substitution works because the function $\frac{z^2}{y}$ is decreasing with respect to y, and $\min(2,\sqrt{x_3})$ is concave, satisfying the composition rule.

I have a doubt regarding this. The composition rule requires that $\frac{z^2}{y}$ is non-increasing in each argument, so why can we ignore the monotonicity with respect to z just because $1^T \cdot x$ is an affine function? It confuses me a lot.

I hope someone knowledgeable can provide a detailed explanation. Thank you very much!

I am unable to arrive at the same conclusion as the professor, regardless of the perspective from which I approach the preservation of convexity.

The video website is https://www.youtube.com/watch?v=U2HRwA7XePU. You can watch this example starting from 29:46.

Answer 1

Listen to the extended composition rule around timestamp 21:39 and define $$h:(y,z)\mapsto \frac{z^2}y, \quad g_1:x\mapsto \min(2,\sqrt x_3), \quad g_2:x\mapsto 1^\top x.$$

$h$ is convex.
$g_1$ is concave and $h$ is decreasing in its first argument.
$g_2$ is affine.

Hence $f:x\mapsto h(g_1(x), g_2(x))$ is convex.